Let 2

Question

Let $\dfrac\pi2 < \theta < \pi$ and $\cot\theta = -\dfrac1{2\sqrt2}$. Then the value of

$$\sin\left(\frac{15\theta}{2}\right)(\cos8\theta+\sin8\theta) + \cos\left(\frac{15\theta}{2}\right)(\cos8\theta-\sin8\theta)$$

is equal to:

Accepted Answer

Expanding the expression:

$$\sin\tfrac{15\theta}{2}\cos8\theta+\sin\tfrac{15\theta}{2}\sin8\theta+\cos\tfrac{15\theta}{2}\cos8\theta-\cos\tfrac{15\theta}{2}\sin8\theta$$

$$= \left[\cos\tfrac{15\theta}{2}\cos8\theta+\sin\tfrac{15\theta}{2}\sin8\theta\right] + \left[\sin\tfrac{15\theta}{2}\cos8\theta-\cos\tfrac{15\theta}{2}\sin8\theta\right]$$

$$= \cos\left(8\theta-\tfrac{15\theta}{2}\right) + \sin\left(\tfrac{15\theta}{2}-8\theta\right) = \cos\tfrac\theta2 - \sin\tfrac\theta2$$

Now with $\cot\theta = -\tfrac1{2\sqrt2}$, i.e. $\tan\theta = -2\sqrt2$, and $\theta$ in the second quadrant ($\sin\theta>0,\cos\theta<0$), a right triangle with opposite $2\sqrt2$ and adjacent $1$ gives hypotenuse $3$, so:

$$\sin\theta = \frac{2\sqrt2}{3}, \qquad \cos\theta = -\frac13$$

Using half-angle formulas ($\theta/2 \in (\tfrac\pi4,\tfrac\pi2)$, so both $\sin\tfrac\theta2,\cos\tfrac\theta2 > 0$):

$$\cos\theta = 1-2\sin^2\tfrac\theta2 \implies -\tfrac13 = 1-2\sin^2\tfrac\theta2 \implies \sin^2\tfrac\theta2 = \tfrac23 \implies \sin\tfrac\theta2 = \sqrt{\tfrac23}$$

$$\cos\theta = 2\cos^2\tfrac\theta2-1 \implies -\tfrac13 = 2\cos^2\tfrac\theta2-1 \implies \cos^2\tfrac\theta2 = \tfrac13 \implies \cos\tfrac\theta2 = \sqrt{\tfrac13}$$

So the expression equals $\sqrt{\tfrac13}-\sqrt{\tfrac23} = \dfrac{1-\sqrt2}{\sqrt3}$.

Solutions · 1